21x^2+182x+280=0

Solution for 21x^2+182x+280=0 equation:

21x^2+182x+280=0

a = 21; b = 182; c = +280;
Δ = b2-4ac
Δ = 1822-4·21·280
Δ = 9604
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9604}=98$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(182)-98}{2*21}=\frac{-280}{42}
=-6+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(182)+98}{2*21}=\frac{-84}{42}
=-2
$